Answer:
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g(t) = 0
And
The differential equation
is linear and homogeneous
Step-by-step explanation:
Given that,
The differential equation is -

![e^{t}y' + (9t - \frac{1}{t^{2} + 81 } )y = 0\\e^{t}y' + (\frac{9t(t^{2} + 81 ) - 1}{t^{2} + 81 } )y = 0\\e^{t}y' + (\frac{9t^{3} + 729t - 1}{t^{2} + 81 } )y = 0\\y' + [\frac{9t^{3} + 729t - 1}{e^{t}(t^{2} + 81) } ]y = 0](https://tex.z-dn.net/?f=e%5E%7Bt%7Dy%27%20%2B%20%289t%20-%20%5Cfrac%7B1%7D%7Bt%5E%7B2%7D%20%2B%2081%20%7D%20%29y%20%3D%200%5C%5Ce%5E%7Bt%7Dy%27%20%2B%20%28%5Cfrac%7B9t%28t%5E%7B2%7D%20%2B%2081%20%29%20-%201%7D%7Bt%5E%7B2%7D%20%2B%2081%20%7D%20%29y%20%3D%200%5C%5Ce%5E%7Bt%7Dy%27%20%2B%20%28%5Cfrac%7B9t%5E%7B3%7D%20%2B%20729t%20%20-%201%7D%7Bt%5E%7B2%7D%20%2B%2081%20%7D%20%29y%20%3D%200%5C%5Cy%27%20%2B%20%5B%5Cfrac%7B9t%5E%7B3%7D%20%2B%20729t%20%20-%201%7D%7Be%5E%7Bt%7D%28t%5E%7B2%7D%20%2B%2081%29%20%7D%20%5Dy%20%3D%200)
By comparing with y′+p(t)y=g(t), we get
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g(t) = 0
And
The differential equation
is linear and homogeneous.
Answer:
The answer is "The threat to internal validity is mortality".
Step-by-step explanation:
It signifies that while in terms of your study, persons were "morning." It usually signifies when people leave the studies. There will be only likely if indeed the experiment lasts a long period that it'll be a substantial danger to validity and reliability because the chance for drops increases. Eight content validity risks are history, maturity, instrument, test, choice, average regress, human engagement, and mortality.
Answer:
Step-by-step explanation:
y = 2x - 1
y = x - 8 ------ (i)
substitute y = 2x - 1 in equation (i)
2x - 1 = x - 8
2x = x - 8 + 1
2x = x - 7
2x - x = -7
x = -7
Substitute x value in equation (i)
y = -7 - 8
y = -15
This area is equal to the sum of a circle with a radius of 5/2 in and a rectangle 5 by 15 in so:
A=πr^2+xy
A=π(2.5)^2+5*15
A=6.25π+75 in^2
A≈94.63 in^2 (to nearest hundredth) so
A≈95 in^2 (to nearest whole square inch)
Answer:
25 ft
Step-by-step explanation:
The area of this landing pad is 1,962.5 ft², which equals πr².
We solve this equation for r²:
1962.5 ft²
r² = ----------------- = 625 ft²
3.14
Taking the square root of both sides, we get:
r = √(625 ft²) = 25 ft
The radius is 25 ft.